Rotational Dynamics Problem - Rod slipping against Block
- Thread starter Tanya Sharma
- Start date
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
35 replies · 8K views
Physics news on Phys.org
Tanya Sharma
- 1,541
- 135
ehild said:Have you got the direction of the acceleration of the COM?
ehild
Yes , I have got the direction of acceleration of COM.I worked with radial and tangential components and found their resultant i.e acom to make 30° with the tangential component .i.e vertically downwards .This showed that the net acceleration of COM was vertically downwards at the moment rod loses contact with the block .The net resultant force should be vertically downwards i.e -ve y-axis .Now N vanishes so horizontal component of Hinge force is zero .
Science Advisor
Homework Helper
- 15,536
- 1,917
Very well! So you got the correct result that the reaction force is mg/4 at the hinge upward.
You would get the same from the x,y components of acceleration. I like this method, as I usually get confused with the radial and tangential components.
The coordinates of the COM of a homogeneous rod with the ends at (x1,y1) and (x2,y2) are Xcom=(x1+x2)/2, Ycom=(y1+y2)/2. (x1,y1)=(0,0), and (x2,y2)=(x,y) here, so Xcom=x/2, Ycom=y/2.
You derived the x component of acceleration already, the same has to be done with the y component.
y=Lsin(θ) y'=Lcos(θ) θ',
y''= L(-sin(θ)θ'2+cos(θ)θ")=L(-sinθω2+cos(θ)ω')
When loosing contact, y"=L(-1/2 ω2+√3/2 ω')
From the condition that the x component of acceleration is zero, you have got that
ω'=-√3 ω2 and ω2=3g/4L, therefore y"=-3g/2,
and Ycom"=-3g/4.
ehild
You would get the same from the x,y components of acceleration. I like this method, as I usually get confused with the radial and tangential components.
The coordinates of the COM of a homogeneous rod with the ends at (x1,y1) and (x2,y2) are Xcom=(x1+x2)/2, Ycom=(y1+y2)/2. (x1,y1)=(0,0), and (x2,y2)=(x,y) here, so Xcom=x/2, Ycom=y/2.
You derived the x component of acceleration already, the same has to be done with the y component.
y=Lsin(θ) y'=Lcos(θ) θ',
y''= L(-sin(θ)θ'2+cos(θ)θ")=L(-sinθω2+cos(θ)ω')
When loosing contact, y"=L(-1/2 ω2+√3/2 ω')
From the condition that the x component of acceleration is zero, you have got that
ω'=-√3 ω2 and ω2=3g/4L, therefore y"=-3g/2,
and Ycom"=-3g/4.
ehild
Tanya Sharma
- 1,541
- 135
ehild...Thank you very much ...
Its a fantastic experience learning from you .You taught me Physics ,Calculus,Polar Coordinates all in one question
Its a fantastic experience learning from you .You taught me Physics ,Calculus,Polar Coordinates all in one question
Science Advisor
Homework Helper
- 15,536
- 1,917
Tanya Sharma said:ehild...Thank you very much ...
Its a fantastic experience learning from you .You taught me Physics ,Calculus,Polar Coordinates all in one question![]()
So you discovered that it was about relation between polar and Cartesian coordinates? I did not want to confuse you by speaking about polar coordinates:
ehild
Kishlay
- 91
- 3
yes process is correct
Last edited:
Similar threads
Sphere Slipping Against Blocks: Solving for the Speed of the Sphere's Center
- Tanya Sharma
- · Replies 7 ·
- Introductory Physics Homework Help
- Replies
- 7
Block slipping against two blocks
- Vibhor
- · Replies 1 ·
- Introductory Physics Homework Help
- Replies
- 1